diff --git a/solution/1000-1099/1062.Longest Repeating Substring/README.md b/solution/1000-1099/1062.Longest Repeating Substring/README.md index 12a7cb773e4c3..aaed274abecbf 100644 --- a/solution/1000-1099/1062.Longest Repeating Substring/README.md +++ b/solution/1000-1099/1062.Longest Repeating Substring/README.md @@ -66,20 +66,21 @@ tags: ### 方法一：动态规划 -定义 $dp[i][j]$ 表示以 $s[i]$ 和 $s[j]$ 结尾的最长重复子串 🔒 的长度。状态转移方程为： +我们定义 $f[i][j]$ 表示以 $s[i]$ 和 $s[j]$ 结尾的最长重复子串的长度，初始时 $f[i][j]=0$。 + +我们在 $[1, n)$ 的区间内枚举 $i$，在 $[0, i)$ 的区间内枚举 $j$，如果 $s[i]=s[j]$，那么有： $$ -dp[i][j]= +f[i][j]= \begin{cases} -dp[i-1][j-1]+1, & i>0 \cap s[i]=s[j] \\ -1, & i=0 \cap s[i]=s[j] \\ -0, & s[i] \neq s[j] +f[i-1][j-1]+1, & j>0 \\ +1, & j=0 \end{cases} $$ -时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。 +我们求出所有 $f[i][j]$ 的最大值即为答案。 -其中 $n$ 为字符串 $s$ 的长度。 +时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。其中 $n$ 为字符串 $s$ 的长度。 相似题目： @@ -93,13 +94,13 @@ $$ class Solution: def longestRepeatingSubstring(self, s: str) -> int: n = len(s) - dp = [[0] * n for _ in range(n)] + f = [[0] * n for _ in range(n)] ans = 0 - for i in range(n): - for j in range(i + 1, n): + for i in range(1, n): + for j in range(i): if s[i] == s[j]: - dp[i][j] = dp[i - 1][j - 1] + 1 if i else 1 - ans = max(ans, dp[i][j]) + f[i][j] = 1 + (f[i - 1][j - 1] if j else 0) + ans = max(ans, f[i][j]) return ans ``` @@ -109,13 +110,13 @@ class Solution: class Solution { public int longestRepeatingSubstring(String s) { int n = s.length(); + int[][] f = new int[n][n]; int ans = 0; - int[][] dp = new int[n][n]; - for (int i = 0; i < n; ++i) { - for (int j = i + 1; j < n; ++j) { + for (int i = 1; i < n; ++i) { + for (int j = 0; j < i; ++j) { if (s.charAt(i) == s.charAt(j)) { - dp[i][j] = i > 0 ? dp[i - 1][j - 1] + 1 : 1; - ans = Math.max(ans, dp[i][j]); + f[i][j] = 1 + (j > 0 ? f[i - 1][j - 1] : 0); + ans = Math.max(ans, f[i][j]); } } } @@ -130,14 +131,15 @@ class Solution { class Solution { public: int longestRepeatingSubstring(string s) { - int n = s.size(); - vector> dp(n, vector(n)); + int n = s.length(); + int f[n][n]; + memset(f, 0, sizeof(f)); int ans = 0; - for (int i = 0; i < n; ++i) { - for (int j = i + 1; j < n; ++j) { + for (int i = 1; i < n; ++i) { + for (int j = 0; j < i; ++j) { if (s[i] == s[j]) { - dp[i][j] = i ? dp[i - 1][j - 1] + 1 : 1; - ans = max(ans, dp[i][j]); + f[i][j] = 1 + (j > 0 ? f[i - 1][j - 1] : 0); + ans = max(ans, f[i][j]); } } } @@ -149,26 +151,66 @@ public: #### Go ```go -func longestRepeatingSubstring(s string) int { +func longestRepeatingSubstring(s string) (ans int) { n := len(s) - dp := make([][]int, n) - for i := range dp { - dp[i] = make([]int, n) + f := make([][]int, n) + for i := range f { + f[i] = make([]int, n) } - ans := 0 - for i := 0; i < n; i++ { - for j := i + 1; j < n; j++ { + for i := 1; i < n; i++ { + for j := 0; j < i; j++ { if s[i] == s[j] { - if i == 0 { - dp[i][j] = 1 - } else { - dp[i][j] = dp[i-1][j-1] + 1 + if j > 0 { + f[i][j] = f[i-1][j-1] } - ans = max(ans, dp[i][j]) + f[i][j]++ + ans = max(ans, f[i][j]) } } } - return ans + return +} +``` + +#### TypeScript + +```ts +function longestRepeatingSubstring(s: string): number { + const n = s.length; + const f: number[][] = Array.from({ length: n }).map(() => Array(n).fill(0)); + let ans = 0; + for (let i = 1; i < n; ++i) { + for (let j = 0; j < i; ++j) { + if (s[i] === s[j]) { + f[i][j] = 1 + (f[i - 1][j - 1] || 0); + ans = Math.max(ans, f[i][j]); + } + } + } + return ans; +} +``` + +#### Rust + +```rust +impl Solution { + pub fn longest_repeating_substring(s: String) -> i32 { + let n = s.len(); + let mut f = vec![vec![0; n]; n]; + let mut ans = 0; + let s = s.as_bytes(); + + for i in 1..n { + for j in 0..i { + if s[i] == s[j] { + f[i][j] = if j > 0 { f[i - 1][j - 1] + 1 } else { 1 }; + ans = ans.max(f[i][j]); + } + } + } + ans + } } ``` diff --git a/solution/1000-1099/1062.Longest Repeating Substring/README_EN.md b/solution/1000-1099/1062.Longest Repeating Substring/README_EN.md index c6f581a644da9..9355ab3aadcaf 100644 --- a/solution/1000-1099/1062.Longest Repeating Substring/README_EN.md +++ b/solution/1000-1099/1062.Longest Repeating Substring/README_EN.md @@ -62,7 +62,27 @@ tags: -### Solution 1 +### Solution 1: Dynamic Programming + +We define $f[i][j]$ to represent the length of the longest repeating substring ending with $s[i]$ and $s[j]$. Initially, $f[i][j]=0$. + +We enumerate $i$ in the range $[1, n)$ and enumerate $j$ in the range $[0, i)$. If $s[i]=s[j]$, then we have: + +$$ +f[i][j]= +\begin{cases} +f[i-1][j-1]+1, & j>0 \\ +1, & j=0 +\end{cases} +$$ + +The answer is the maximum value of all $f[i][j]$. + +The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Where $n$ is the length of the string $s$. + +Similar problems: + +- [1044. Longest Duplicate Substring 🔒](https://github.com/doocs/leetcode/blob/main/solution/1000-1099/1044.Longest%20Duplicate%20Substring/README_EN.md) @@ -72,13 +92,13 @@ tags: class Solution: def longestRepeatingSubstring(self, s: str) -> int: n = len(s) - dp = [[0] * n for _ in range(n)] + f = [[0] * n for _ in range(n)] ans = 0 - for i in range(n): - for j in range(i + 1, n): + for i in range(1, n): + for j in range(i): if s[i] == s[j]: - dp[i][j] = dp[i - 1][j - 1] + 1 if i else 1 - ans = max(ans, dp[i][j]) + f[i][j] = 1 + (f[i - 1][j - 1] if j else 0) + ans = max(ans, f[i][j]) return ans ``` @@ -88,13 +108,13 @@ class Solution: class Solution { public int longestRepeatingSubstring(String s) { int n = s.length(); + int[][] f = new int[n][n]; int ans = 0; - int[][] dp = new int[n][n]; - for (int i = 0; i < n; ++i) { - for (int j = i + 1; j < n; ++j) { + for (int i = 1; i < n; ++i) { + for (int j = 0; j < i; ++j) { if (s.charAt(i) == s.charAt(j)) { - dp[i][j] = i > 0 ? dp[i - 1][j - 1] + 1 : 1; - ans = Math.max(ans, dp[i][j]); + f[i][j] = 1 + (j > 0 ? f[i - 1][j - 1] : 0); + ans = Math.max(ans, f[i][j]); } } } @@ -109,14 +129,15 @@ class Solution { class Solution { public: int longestRepeatingSubstring(string s) { - int n = s.size(); - vector> dp(n, vector(n)); + int n = s.length(); + int f[n][n]; + memset(f, 0, sizeof(f)); int ans = 0; - for (int i = 0; i < n; ++i) { - for (int j = i + 1; j < n; ++j) { + for (int i = 1; i < n; ++i) { + for (int j = 0; j < i; ++j) { if (s[i] == s[j]) { - dp[i][j] = i ? dp[i - 1][j - 1] + 1 : 1; - ans = max(ans, dp[i][j]); + f[i][j] = 1 + (j > 0 ? f[i - 1][j - 1] : 0); + ans = max(ans, f[i][j]); } } } @@ -128,26 +149,66 @@ public: #### Go ```go -func longestRepeatingSubstring(s string) int { +func longestRepeatingSubstring(s string) (ans int) { n := len(s) - dp := make([][]int, n) - for i := range dp { - dp[i] = make([]int, n) + f := make([][]int, n) + for i := range f { + f[i] = make([]int, n) } - ans := 0 - for i := 0; i < n; i++ { - for j := i + 1; j < n; j++ { + for i := 1; i < n; i++ { + for j := 0; j < i; j++ { if s[i] == s[j] { - if i == 0 { - dp[i][j] = 1 - } else { - dp[i][j] = dp[i-1][j-1] + 1 + if j > 0 { + f[i][j] = f[i-1][j-1] } - ans = max(ans, dp[i][j]) + f[i][j]++ + ans = max(ans, f[i][j]) } } } - return ans + return +} +``` + +#### TypeScript + +```ts +function longestRepeatingSubstring(s: string): number { + const n = s.length; + const f: number[][] = Array.from({ length: n }).map(() => Array(n).fill(0)); + let ans = 0; + for (let i = 1; i < n; ++i) { + for (let j = 0; j < i; ++j) { + if (s[i] === s[j]) { + f[i][j] = 1 + (f[i - 1][j - 1] || 0); + ans = Math.max(ans, f[i][j]); + } + } + } + return ans; +} +``` + +#### Rust + +```rust +impl Solution { + pub fn longest_repeating_substring(s: String) -> i32 { + let n = s.len(); + let mut f = vec![vec![0; n]; n]; + let mut ans = 0; + let s = s.as_bytes(); + + for i in 1..n { + for j in 0..i { + if s[i] == s[j] { + f[i][j] = if j > 0 { f[i - 1][j - 1] + 1 } else { 1 }; + ans = ans.max(f[i][j]); + } + } + } + ans + } } ``` diff --git a/solution/1000-1099/1062.Longest Repeating Substring/Solution.cpp b/solution/1000-1099/1062.Longest Repeating Substring/Solution.cpp index e8d4237e5ef1b..4e7811621fec3 100644 --- a/solution/1000-1099/1062.Longest Repeating Substring/Solution.cpp +++ b/solution/1000-1099/1062.Longest Repeating Substring/Solution.cpp @@ -1,17 +1,18 @@ class Solution { public: int longestRepeatingSubstring(string s) { - int n = s.size(); - vector> dp(n, vector(n)); + int n = s.length(); + int f[n][n]; + memset(f, 0, sizeof(f)); int ans = 0; - for (int i = 0; i < n; ++i) { - for (int j = i + 1; j < n; ++j) { + for (int i = 1; i < n; ++i) { + for (int j = 0; j < i; ++j) { if (s[i] == s[j]) { - dp[i][j] = i ? dp[i - 1][j - 1] + 1 : 1; - ans = max(ans, dp[i][j]); + f[i][j] = 1 + (j > 0 ? f[i - 1][j - 1] : 0); + ans = max(ans, f[i][j]); } } } return ans; } -}; \ No newline at end of file +}; diff --git a/solution/1000-1099/1062.Longest Repeating Substring/Solution.go b/solution/1000-1099/1062.Longest Repeating Substring/Solution.go index d1ae5a55276d0..a4371e26d8962 100644 --- a/solution/1000-1099/1062.Longest Repeating Substring/Solution.go +++ b/solution/1000-1099/1062.Longest Repeating Substring/Solution.go @@ -1,21 +1,19 @@ -func longestRepeatingSubstring(s string) int { +func longestRepeatingSubstring(s string) (ans int) { n := len(s) - dp := make([][]int, n) - for i := range dp { - dp[i] = make([]int, n) + f := make([][]int, n) + for i := range f { + f[i] = make([]int, n) } - ans := 0 - for i := 0; i < n; i++ { - for j := i + 1; j < n; j++ { + for i := 1; i < n; i++ { + for j := 0; j < i; j++ { if s[i] == s[j] { - if i == 0 { - dp[i][j] = 1 - } else { - dp[i][j] = dp[i-1][j-1] + 1 + if j > 0 { + f[i][j] = f[i-1][j-1] } - ans = max(ans, dp[i][j]) + f[i][j]++ + ans = max(ans, f[i][j]) } } } - return ans -} \ No newline at end of file + return +} diff --git a/solution/1000-1099/1062.Longest Repeating Substring/Solution.java b/solution/1000-1099/1062.Longest Repeating Substring/Solution.java index 4f745bde390e9..fc16f454e3d52 100644 --- a/solution/1000-1099/1062.Longest Repeating Substring/Solution.java +++ b/solution/1000-1099/1062.Longest Repeating Substring/Solution.java @@ -1,16 +1,16 @@ class Solution { public int longestRepeatingSubstring(String s) { int n = s.length(); + int[][] f = new int[n][n]; int ans = 0; - int[][] dp = new int[n][n]; - for (int i = 0; i < n; ++i) { - for (int j = i + 1; j < n; ++j) { + for (int i = 1; i < n; ++i) { + for (int j = 0; j < i; ++j) { if (s.charAt(i) == s.charAt(j)) { - dp[i][j] = i > 0 ? dp[i - 1][j - 1] + 1 : 1; - ans = Math.max(ans, dp[i][j]); + f[i][j] = 1 + (j > 0 ? f[i - 1][j - 1] : 0); + ans = Math.max(ans, f[i][j]); } } } return ans; } -} \ No newline at end of file +} diff --git a/solution/1000-1099/1062.Longest Repeating Substring/Solution.py b/solution/1000-1099/1062.Longest Repeating Substring/Solution.py index a03407f1e5c2a..ee10cc0fad4f1 100644 --- a/solution/1000-1099/1062.Longest Repeating Substring/Solution.py +++ b/solution/1000-1099/1062.Longest Repeating Substring/Solution.py @@ -1,11 +1,11 @@ class Solution: def longestRepeatingSubstring(self, s: str) -> int: n = len(s) - dp = [[0] * n for _ in range(n)] + f = [[0] * n for _ in range(n)] ans = 0 - for i in range(n): - for j in range(i + 1, n): + for i in range(1, n): + for j in range(i): if s[i] == s[j]: - dp[i][j] = dp[i - 1][j - 1] + 1 if i else 1 - ans = max(ans, dp[i][j]) + f[i][j] = 1 + (f[i - 1][j - 1] if j else 0) + ans = max(ans, f[i][j]) return ans diff --git a/solution/1000-1099/1062.Longest Repeating Substring/Solution.rs b/solution/1000-1099/1062.Longest Repeating Substring/Solution.rs new file mode 100644 index 0000000000000..4ed8e53d075a4 --- /dev/null +++ b/solution/1000-1099/1062.Longest Repeating Substring/Solution.rs @@ -0,0 +1,18 @@ +impl Solution { + pub fn longest_repeating_substring(s: String) -> i32 { + let n = s.len(); + let mut f = vec![vec![0; n]; n]; + let mut ans = 0; + let s = s.as_bytes(); + + for i in 1..n { + for j in 0..i { + if s[i] == s[j] { + f[i][j] = if j > 0 { f[i - 1][j - 1] + 1 } else { 1 }; + ans = ans.max(f[i][j]); + } + } + } + ans + } +} diff --git a/solution/1000-1099/1062.Longest Repeating Substring/Solution.ts b/solution/1000-1099/1062.Longest Repeating Substring/Solution.ts new file mode 100644 index 0000000000000..42120294b543e --- /dev/null +++ b/solution/1000-1099/1062.Longest Repeating Substring/Solution.ts @@ -0,0 +1,14 @@ +function longestRepeatingSubstring(s: string): number { + const n = s.length; + const f: number[][] = Array.from({ length: n }).map(() => Array(n).fill(0)); + let ans = 0; + for (let i = 1; i < n; ++i) { + for (let j = 0; j < i; ++j) { + if (s[i] === s[j]) { + f[i][j] = 1 + (f[i - 1][j - 1] || 0); + ans = Math.max(ans, f[i][j]); + } + } + } + return ans; +}